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grab text between two patern

Check the code bellow. I want to grab everything between this id="a-popover-sp-info-popover- until ". I already tried to use following Regex.Match formula but there is syntax error. Its not valid in c#. How can i do this in proper way. My goal is to grab ABC123 text.
string foo = #id="a-popover-sp-info-popover-ABC123";
string output = Regex.Match(foo, #"id="a-popover-sp-info-popover-(.*)"").Groups[1].Value;
i need to grab only text: ABC123

since your pattern is so rigid, actually the string.Split method could also do the trick:
string output1 = foo.Split(new string[] {"info-popover-"}, StringSplitOptions.RemoveEmptyEntries)
.Last()
.TrimEnd('"');
Console.WriteLine(output1);
Output:
ABC123

You have to make sure to surround your strings with quotation marks ".
If you want to have quotation marks inside of your string you have to escape them with a backslash:
string foo = "id=\"a-popover-sp-info-popover-ABC123\"";
string output = Regex.Match(foo, "id=\"a-popover-sp-info-popover-(.*)\"").Groups[1].Value;

string pattern = "id=\"a-popover-sp-info-popover-[A-Z]{3}[1-9]{3}\"";
string input = "id=\"a-popover-sp-info-popover-ABC123\"";
Match m = Regex.Match(input, pattern);
if (m.Success) Console.WriteLine("Found '{0}'", m.Value);

How do I replace all instances of any special characters between each occurrence of a set of delimiters in a string?

I'm attempting to replace all instances of any special characters between each occurrence of a set of delimiters in a string. I believe the solution will include some combination of a regular expression match to retrieve the text between each set of delimiters and a regular expression replace to replace each offending character within the match with a space. Here’s what I have so far:
string input = "***XX*123456789~N3*123 E. Fake St. Apt# 456~N4*Beverly Hills*CA*902122405~REF*EI*902122405~HL*1*1*50*0~SBR*P*18*******MA~NM1*IL*1*Tom*Thompson*T***MI*123456789A~N3*456 W. False Ave.*Apt. #6B~N4*Beverly Hills*CA*90210~DMG*";
string matchPattern = "(~N3\\*)(.*?)(~N4\\*)";
string replacePattern = "[^0-9a-zA-Z ]?";
var matches = Regex.Matches(input, matchPattern);
foreach (Match match in matches)
{
match.Value = "~N3*" + Regex.Replace(match.Value, replacePattern, " ") + "~N4*";
}
MessageBox.Show(input);
I would expect the message box to show the following:
"***XX*123456789~N3*123 E Fake St Apt 456~N4*Beverly Hills*CA*902122405~REF*EI*902122405~HL*1*1*50*0~SBR*P*18*******MA~NM1*IL*1*Tom*Thompson*T***MI*123456789A~N3*456 W False Ave *Apt 6B~N4*Beverly Hills*CA*90210~DMG*"
Obviously this isn’t working because I can’t assign to the matched value inside the loop, but I hope you can follow my thought process. It is important that any characters which are not between the delimiters remain unchanged. Any direction or advice would be helpful. Thank you so much!

Use a Regex.Replace with a match evaluator where you may call the second Regex.Replace:
string input = "***XX*123456789~N3*123 E. Fake St. Apt# 456~N4*Beverly Hills*CA*902122405~REF*EI*902122405~HL*1*1*50*0~SBR*P*18*******MA~NM1*IL*1*Tom*Thompson*T***MI*123456789A~N3*456 W. False Ave.*Apt. #6B~N4*Beverly Hills*CA*90210~DMG*";
string matchPattern = #"(~N3\*)(.*?)(~N4\*)";
string replacePattern = "[^0-9a-zA-Z ]";
string res = Regex.Replace(input, matchPattern, m =>
string.Format("{0}{1}{2}",
m.Groups[1].Value,
Regex.Replace(m.Groups[2].Value, replacePattern, " "), // Here, you modify just inside the 1st regex matches
m.Groups[3].Value));
Console.Write(res); // Just to print the demo result
// => ***XX*123456789~N3*123 E Fake St Apt 456~N4*Beverly Hills*CA*902122405~REF*EI*902122405~HL*1*1*50*0~SBR*P*18*******MA~NM1*IL*1*Tom*Thompson*T***MI*123456789A~N3*456 W False Ave Apt 6B~N4*Beverly Hills*CA*90210~DMG*
See the C# demo
Actually, since ~N3* and ~N4* are literal strings, you may use a single capturing group in the pattern and then add those delimiters as hard-coded in the match evaluator, but it is up to you to decide what suits you best.

Regex removing empty spaces when using replace

My situation is not about removing empty spaces, but keeping them. I have this string >[database values] which I would like to find. I created this RegEx to find it then go in and remove the >, [, ]. The code below takes a string that is from a document. The first pattern looks for anything that is surrounded by >[some stuff] it then goes in and "removes" >, [, ]
string decoded = "document in string format";
string pattern = #">\[[A-z, /, \s]*\]";
string pattern2 = #"[>, \[, \]]";
Regex rgx = new Regex(pattern);
Regex rgx2 = new Regex(pattern2);
foreach (Match match in rgx.Matches(decoded))
{
string replacedValue= rgx2.Replace(match.Value, "");
Console.WriteLine(match.Value);
Console.WriteLine(replacedValue);
What I am getting in first my Console.WriteLine is correct. So I would be getting things like >[123 sesame St]. But my second output shows that my replace removes not just the characters but the spaces so I would get something like this 123sesameSt. I don't see any space being replaced in my Regex. Am I forgetting something, perhaps it is implicitly in a replace?

The [A-z, /, \s] and [>, \[, \]] in your patterns are also looking for commas and spaces. Just list the characters without delimiting them, like this: [A-Za-z/\s]
string pattern = #">\[[A-Za-z/\s]*\]";
string pattern2 = #"[>,\[\]]";
Edit to include Casimir's tip.

After rereading your question (if I understand well) I realize that your two steps approach is useless. You only need one replacement using a capture group:
string pattern = #">\[([^]]*)]";
Regex rgx = new Regex(pattern);
string result = rgx.Replace(yourtext, "$1");
pattern details:
>\[ # literals: >[
( # open the capture group 1
[^]]* # all that is not a ]
) # close the capture group 1
] # literal ]
the replacement string refers to the capture group 1 with $1

By defining [>, \[, \]] in pattern2 you define a character group consisting of single characters like >, ,, , [ and every other character you listed in the square brackets. But I guess you don't want to match space and ,. So if you don't want to match them leave them out like
string pattern2 = #"[>\[\]]";
Alternatively, you could use
string pattern2 = #"(>\[|\])";
Thereby, you either match >[ or ] which better expresses your intention.

Regex - Find from both sides only if it has spaces

I need some help on Regex. I need to find a word that is surrounded by whatever element, for example - *. But I need to match it only if it has spaces or nothing on the ether sides. For example if it is at start of the text I can't really have space there, same for end.
Here is what I came up to
string myString = "You will find *me*, and *me* also!";
string findString = #"(\*(.*?)\*)";
string foundText;
MatchCollection matchCollection = Regex.Matches(myString, findString);
foreach (Match match in matchCollection)
{
foundText = match.Value.Replace("*", "");
myString = myString.Replace(match.Value, "->" + foundText + "<-");
match.NextMatch();
}
Console.WriteLine(myString);
You will find ->me<-, and ->me<- also!
Works correct, the problem is when I add * in the middle of text, I don't want it to match then.
Example: You will find *m*e*, and *me* also!
Output: You will find ->m<-e->, and <-me* also!
How can I fix that?

Try the following pattern:
string findString = #"(?<=\s|^)\*(.*?)\*(?=\s|$)";
(?<=\s|^)X will match any X only if preceded by a space-char (\s), or the start-of-input, and
X(?=\s|$) matches any X if followed by a space-char (\s), or the end-of-input.
Note that it will not match *me* in foo *me*, bar since the second * has a , after it! If you want to match that too, you need to include the comma like this:
string findString = #"(?<=[\s,]|^)\*(.*?)\*(?=[\s,]|$)";
You'll need to expand the set [\s,] as you see fit, of course. You might want to add !, ? and . at the very least: [\s,!?.] (and no, . and ? do not need to be escaped inside a character-set!).
EDIT
A small demo:
string Txt = "foo *m*e*, bar";
string Pattern = #"(?<=[\s,]|^)\*(.*?)\*(?=[\s,]|$)";
Console.WriteLine(Regex.Replace(Txt, Pattern, ">$1<"));
which would print:
>m*e<

You can add "beginning of line or space" and "space or end of line" around your match:
(^|\s)\*(.*?)\*(\s|$)
You'll now need to refer to the middle capture group for the match string.

Regular expression: take string literally with special re-characters

Maybe simple question..
String text = "fake 43 60 fake";
String patt = "[43.60]";
Match m = Regex.Match(text, patt)
In this situation, m.Success = true because the dot replace any character (also the space). But I must match the string literally in the patt.
Of course, I can use the '\' before the dot in the patt
String patt = #"[43\.60]";
So the m.Success = false, but there's more special characters in the Regular Expression-world.
My question is, how can I use regular expression that a string will be literally taken as it set. So '43.60' must be match with exactly '43.60'. '43?60' must be match with '43?60'....
thanks.

To get a regex-safe literal:
string escaped = Regex.Escape(input);
For example, to match the literal [43.60]:
string escaped = Regex.Escape(#"[43.60]");
gives the string with content: \[43\.60].
You can then use this escaped content to create a regex; for example:
string find = "43?60";
string escaped = Regex.Escape(find);
bool match = Regex.IsMatch("abc 43?60", escaped);
Note that in many cases you will want to combine the escaped string with some other regex fragment to make a complete pattern.